Chapter 14.5, Problem 20WE

To determine

To find: the required roots in $\text{x+iy}$ form.

Answer to Problem 20WE

  ${\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left({\text{cos60}}^{\text{0}}{\text{+isin60}}^{\text{0}}\right)\text{=}{\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left(\frac{\text{1}}{\text{2}}\text{+i}\frac{\sqrt{3}}{\text{2}}\right)$ .

  ${\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left({\text{cos150}}^{\text{0}}{\text{+isin150}}^{\text{0}}\right)\text{=}{\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left(\text{-}\frac{\sqrt{3}}{\text{2}}\text{+i}\frac{\text{1}}{\text{2}}\right)$ .

  ${\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left({\text{cos240}}^{\text{0}}{\text{+isin240}}^{\text{0}}\right)\text{=}{\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left(-\frac{\text{1}}{\text{2}}\text{-i}\frac{\sqrt{\text{3}}}{\text{2}}\right)$ .

  ${\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left({\text{cos330}}^{\text{0}}{\text{+isin330}}^{\text{0}}\right)\text{=}{\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left(\frac{1}{\text{2}}\text{-i}\frac{\sqrt{3}}{\text{2}}\right)$ .

Explanation of Solution

Given:

  $\text{-6-6i}\sqrt{\text{3}}$ .

Concept used:

If $\text{x}$ is the an $\text{nth}$ root of unity, then it satisfies the relation:

  ${\text{x}}^{\text{n}}\text{=1}$ .

  $\text{=cos0+isin0}$ .

  ${\text{x}}^{\text{n}}\text{=cos2kπ+isin2kπ}$ where $\text{k}$ is an integer.

Taking $\text{nth}$ root on both sides:

  ${\left({\text{x}}^{\text{n}}\right)}^{\frac{\text{1}}{\text{n}}}\text{=}{\left(\text{cos2kπ+isin2kπ}\right)}^{\frac{\text{1}}{\text{n}}}$ .

  $\text{x=}{\left(\text{cos2kπ+isin2kπ}\right)}^{\frac{\text{1}}{\text{n}}}$ .

According $\text{de moivre}$ s law $\text{x=}{\left(\text{cos2kπ+isin2kπ}\right)}^{\frac{\text{1}}{\text{n}}}$ can be written as:

In radian: $\text{x=}\left(\text{cos}\frac{\text{2kπ}}{\text{n}}\text{+isin}\frac{\text{2kπ}}{\text{n}}\right)$ .

In degree: $\text{x=}\left(\text{cos}\frac{\text{360k}}{\text{n}}\text{+isin}\frac{\text{360k}}{\text{n}}\right)$ .

Calculation:

The fourth roots of $\text{-6-6i}\sqrt{\text{3}}$ is:

Equation in polar form:

Here from the given equation $\text{x=0}$ and $\text{y=i}$ .

  $\text{z=x+iy=rcosθ+irsinθ}$ .

  $\text{r=}\sqrt{{\left(-6\right)}^{\text{2}}\text{+}{\left(\text{-6}\sqrt{3}\right)}^{\text{2}}}=12$ .

  $\text{r=}12$ .

  $\text{x=rcosθ}$ and $\text{y=rsinθ}$ .

  $\text{cosθ=}\frac{\text{x}}{\text{r}}$ and $\text{sinθ=}\frac{\text{y}}{\text{r}}$ .

  $\text{cosθ=}\frac{\text{x}}{\text{r}}=-\frac{6}{12}=-\frac{1}{2}$ .

  $\text{sinθ=}\frac{\text{y}}{\text{r}}\text{=}\frac{\text{-6}\sqrt{\text{3}}}{\text{12}}\text{=-}\frac{\sqrt{\text{3}}}{\text{2}}$ .

From unit circle the angle can be calculated as:

  ${\text{θ=240}}^0$ .

  ${\text{z}}^4{\text{=r}}^4\left(\text{cos4θ+isin4θ}\right)=12\left({\text{cos240}}^0{\text{+isin240}}^0\right)$

  $\text{r=}{\left(\text{12}\right)}^{\frac{1}{4}}{\text{and cos4θ=240}}^{\text{0}}{\text{,sin4θ=240}}^{\text{0}}$ .

  ${\text{4θ=240}}^{\text{0}}\text{+k}{\text{.360}}^{\text{0}}$ .

  ${\text{θ=60}}^{\text{0}}\text{+k}{\text{.90}}^{\text{0}}$

For $\text{k=0,1,2 and 3}$ .

  ${\text{θ=60}}^{\text{0}}{\text{,150}}^{\text{0}}\text{,}24{\text{0}}^{\text{0}}{\text{ and 330}}^0$ .

Hence three cube roots are:

  ${\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left({\text{cos60}}^{\text{0}}{\text{+isin60}}^{\text{0}}\right)\text{=}{\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left(\frac{\text{1}}{\text{2}}\text{+i}\frac{\sqrt{3}}{\text{2}}\right)$ .

  ${\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left({\text{cos150}}^{\text{0}}{\text{+isin150}}^{\text{0}}\right)\text{=}{\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left(\text{-}\frac{\sqrt{3}}{\text{2}}\text{+i}\frac{\text{1}}{\text{2}}\right)$ .

  ${\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left({\text{cos240}}^{\text{0}}{\text{+isin240}}^{\text{0}}\right)\text{=}{\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left(-\frac{\text{1}}{\text{2}}\text{-i}\frac{\sqrt{\text{3}}}{\text{2}}\right)$ .

  ${\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left({\text{cos330}}^{\text{0}}{\text{+isin330}}^{\text{0}}\right)\text{=}{\left(\text{12}\right)}^{\frac{\text{1}}{\text{4}}}\left(\frac{1}{\text{2}}\text{-i}\frac{\sqrt{3}}{\text{2}}\right)$ .